Hall Effect: Hall coefficient, Hall potential, Hall Voltage


What is Hall Effect? Derive an expression for Hall coefficient, Hall potential, Hall Voltage and Hall angle. What are the importance of the Hall effect?

Answer:

Hall Effect

The Hall Effect discovered by Edwin Hall in 1879. He state that “when we flow a current along the length of the conducting  ribbon (conductor or semiconductor) and placed it into the transverse magnetic field, then an electric potential developed across the ribbon. The direction of electric potential is perpendicular to applied current I and magnetic field B. The electric potential is known as Hall potential and generated field is known as Hall Field and this effect is known as “Hall Effect”.

Hall Effect

If we take a conducting ribbon flowing current in +ve X-direction and the magnetic field is applied in the +ve Z-direction. A force is experienced on charge carrier in the downward direction of conducting ribbon means –ve Y- direction. Through this force the +ve charges arranged at upper edegs and –ve charges arranged at bottom edges of the ribbon. The separation of charge carrier developed the electric field inside the ribbon in the direction of Y.

If we consider a ribbon has a width W and thickness is T. There are two forces acts on electron due to electric field and magnetic field

i.e.

Electric field force on electron = -eEH                                                                                  (1)

Magnetic field force on electron = -eVxBZ                                                                                 (2)

 

In equilibrium condition,   -eEH = -eVxBZ

EH = VxBZ                                                                                                                (3)

Now we define current density (Current density is a measure of the density of an electric current.) in the direction of X i.e.

Jx = -neVx      ( n = no. of charge carrier / unit Volume)

Vx = – Jx / ne                                                              (4)

Fron equation (4) and (3)

EH = -BZ Jx / ne = -(1/ne) BZ Jx                                  (5)

RH = -(1/ne)                                             (6)

RH  is known as Hall Coefficient.

EH = RH BZ Jx

RH = EH / BZ Jx                                                                              (7)

If we take length L and breadth b of the conducting ribbon and hall field EH and developed potential is VH

Then EH = VH / L                                                     (8)

RH = VH / L BZ Jx

VH = RH L BZ Jx                                                                             (9)

Again current density is equal to the current per unit area

i.e Jx  = Ix/ LB

VH = RH L BZ Ix/ LB

VH = RH BZ Ix/ B                          (10)

VH  is known as Hall Potential.

Calculation of Hall angle and Mobility of charge carrier:

The charge carrier mobility is equal to the drift velocity per unit electric field

i.e. µ = Vx/ E

Since for equation (3) EH = VxBZ  

then EH = µEBZ

EH = µEBZ

Since EH = RH BZ Jx

Compare both equation µEBZ  = RH BZ Jx

µ= RH Jx/ E                                                                                        (11)

Mobility µ= sRH                              (s = Jx/ E current density per unit electric field)

Since EH = µEBZ

µ= EH  / EBZ

qH = EH  / E is known as Hall angle.

 

The importance of the Hall effect is supported by the need to determine accurately

  • carrier density.
  • electrical resistivity.
  • and the mobility of carriers in semiconductors.
  • Hall Effect proved that electrons are the majority carriers in all the metals and n-type semiconductors.
  • In p-type semiconductors, holes are the majority carriers.

 

Problem 1: Find Hall coefficient for 5 x 1028 atom / m3 in copper block.

Solution: RH =-1/ne

RH = -1/5 x 1028 x 1.6 x 10-19

RH = -0.125 x 10-9 m3/C

Problem 2: Calculate mobility and charge carrier density when the resistivity of doped Si sample is 9 x 10-3 Ω–m and the hall coefficient is 3.6 x 10-4 m3/C.

Solution: Given that  r =9 x 10-3 Ω–m; RH = 3.6 x  10-4 m3/C

So conductivity s = 1 / r = 1/ 9 x 10-3 /Ω–m

s = 111.1 / Ω–m

r = ne;

n = r/e = 1/ RHe

n = 1/ 1.6 x 10-4 x 1.6 x 10-19

n = 1.7363 x 1022 / m3

Mobility µ = RH s

µ = 111.1 x 3.6 x  10-4

µ = 0.04 m2/V-s

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